440 Sharp solar panels Panel rating = 235 W Overall derating factor = 85% Insolation, Is = 4.8 kWh/(m2 day) Utility Electricity Cost = $0.16/kWh Total Project Cost = $1.2 million Panel Dimensions: Length = 1.64 m, Width = 0.994 m A. What is the total rated capacity (kW) of this array? B. How much electricity (kWh/yr)can this array generate in an average year? C. If the project cost $1.2 million and the cost of electricity is $0.16/kWh, what is the simple payback time (years)? D. What is the approximate efficiency (%) for the PV panels (not considering the derating factor)?
440 Sharp solar panels Panel rating = 235 W Overall derating factor = 85% Insolation, Is = 4.8 kWh/(m2 day) Utility Electricity Cost = $0.16/kWh Total Project Cost = $1.2 million Panel Dimensions: Length = 1.64 m, Width = 0.994 m A. What is the total rated capacity (kW) of this array? B. How much electricity (kWh/yr)can this array generate in an average year? C. If the project cost $1.2 million and the cost of electricity is $0.16/kWh, what is the simple payback time (years)? D. What is the approximate efficiency (%) for the PV panels (not considering the derating factor)?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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- 440 Sharp solar panels
- Panel rating = 235 W
- Overall derating factor = 85%
- Insolation, Is = 4.8 kWh/(m2 day)
- Utility Electricity Cost = $0.16/kWh
- Total Project Cost = $1.2 million
- Panel Dimensions: Length = 1.64 m, Width = 0.994 m
A. What is the total rated capacity (kW) of this array?
B. How much electricity (kWh/yr)can this array generate in an average year?
C. If the project cost $1.2 million and the cost of electricity is $0.16/kWh, what is the simple payback time (years)?
D. What is the approximate efficiency (%) for the PV panels (not considering the derating factor)?
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